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Cubing the square (Posted on 2013-01-30) Difficulty: 3 of 5

I'm thinking of a number, x, whose last 3 digits are somewhere between 400 and 500.

You could probably guess from those digits that it might be a square or a cube, but in fact it turns out to be both, and in fact the smallest possible such number.

What are the prime factors of x?

Extra credit: feeling adventurous, I next computed the smallest number, y, having the same last 3 digits, that was a seventh power, as well as a square and a cube.

What are the prime factors of y?

See The Solution Submitted by broll    
Rating: 5.0000 (1 votes)

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Solution found y, got bonus | Comment 2 of 8 |
We seek n^42=y to end in 441

Let n be of the form 10a + b

n^42 = (10a + b)^42
the last few terms of this binomial expansion are
11480*10^3*a^3*b^39 + [this term doesn't contribute to the last 3 digits]
861*10^2*a^2*b^40 +
42*10*a*b^41 +

Only the last term contributes to the last digit.  b^42 only ends in 1 if b=1.
The second and third to the end terms are then
86100a^2 + 420a
to make the third and second digits from the end be 44 we need
10a^2 + 42a to end in 44

A quick search finds a=12
so n = 10a+b = 121

y = 121^42 = 2999062754174580621444557664122388274963717433652437940577512878446736163557021296243441
= 11^84 = (11^42)^2 = (11^28)^3 = (11^12)^7

Since 121 = 11^2 the only prime factor of y is 11

  Posted by Jer on 2013-01-30 14:51:54
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